Question with virtual functions

Question

I have two classes:

class x {
public:
  virtual void hello() {
    std::cout << "x" << std::endl;
  }
};

class y : public x {
public:
  void hello() {
    std::cout << "y" << std::endl;
  }
};

Can someone explain why the following two calls to hello() print different messages? Why don't they both print "y"? Is it because the first one is a copy while the second one actually points to the object in memory?

int main() {
  y  a;

  x b = a;
  b.hello(); // prints x

  x* c = &a;
  c->hello(); // prints y
  return 0;
}

I think this is the first time I've seen someone correctly guess the cause, good job. — GManNickG
– GManNickG, Commented Jan 28, 2011 at 7:02

StuartLC · Accepted Answer · 2011-01-28 07:01:24Z

6

Yes, you are right

x b = a;

Invokes a copy constructor (b IS an 'x')

x& b = a;

Assigns a reference and will use the override (b is still actually a 'y')

answered Jan 28, 2011 at 7:01

StuartLC

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Community · Accepted Answer · 2017-05-23 12:18:41Z

6

Because x b = a; slices the object.

When this code executes, it creates a new x, not a y, which is a copy of the original y, a'.

edited May 23, 2017 at 12:18

CommunityBot

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answered Jan 28, 2011 at 7:02

John Dibling

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Jonathan Wood · Accepted Answer · 2011-01-28 07:02:11Z

0

x b = a copies a to b. Since b is type x, you end up with an object of type x. An object of type x will print x.

The only way you get y is when you are calling into an object of type y.

answered Jan 28, 2011 at 7:02

Jonathan Wood

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Karmastan · Accepted Answer · 2011-01-28 07:04:28Z

0

b.hello() prints "x" because b is an instance of class X. c->hello() prints "y" because c points to a, and a is an instance of class Y.

What might be confusing for you is that when you write x b = a;, you're creating a new object b and initializing it with a. When you write x* c = &a;, c is not a new object. You just introduced an alias to an existing object.

answered Jan 28, 2011 at 7:04

Karmastan

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